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Question 177

Which Snowflake function will interpret an input string as a JSON document, and produce a VARIANT value?

    Correct Answer: A

    The Snowflake function that interprets an input string as a JSON document and produces a VARIANT value is parse_json(). This function takes a string containing JSON data and converts it into a VARIANT data type, which can store semi-structured data.

Discussion
tejarunOption: A

https://docs.snowflake.com/en/sql-reference/functions/parse_json.html

paknadeemOption: A

The correct answer is A

SV1122

Was on exam Dec 31st 2022

KeshavaMugulurOption: A

This should A for sure

Mallikharjuna452Option: A

parse_json()

pranaligOption: A

Correct Answer: A

AnireddySaikiranReddyOption: B

Correct Answer: B

_yyuktaOption: A

A parse_json()

Marge23Option: A

correct

singhksOption: A

A is the answer. https://docs.snowflake.com/en/sql-reference/functions/parse_json

examtopics_strataOption: A

A is correct

qaiserlatifOption: A

Interprets an input string as a JSON document, producing a VARIANT value. Syntax PARSE_JSON( <expr> )

KarBiswaOption: A

parse_json

ViniJsrOption: A

The correct answer is A.